Let x be a positive element in the spatial tensor product of two non unital C* algebras A and B. Is there a single element $a \otimes b \geq x$? How can we noncommutativize the following proof, in the commutative case: Let $F^2$ be a positive function on $X\times Y$. Define $f(x)=\sup_{y\in Y} F(x,y)$ and $g(y)=\sup_{x\in X} F(x,y)$, then $F^2 \leq fg$.
Yes. For a selfadjoint element $y$, denote $s(y)=\sup{\rm Sp}(y)$. Then for $\gamma \geq s(y)$, one has $$\inf \lbrace s(y  \gamma(e\otimes f)) : 0\le e\le 1,\ 0\le f\le 1\rbrace \le 0.$$ Indeed, if $e_n$ and $f_n$ are approximate units, then so is $g_n:=e_n\otimes f_n$ and $y  \gamma g_n \le yg_n^{1/2} y g_n^{1/2} \to 0$. Now let $y_0 := x \le 1$ and find $0 \le e_n \le 1$ and $0 \le f_n \le 1$ recursively so that the elements $y_{n+1} := y_n  4^{n}(e_n \otimes f_n)$ satisfy $s(y_{n}) \le 4^{n}$ for all $n$. Let $e = \sum_{n=0}^\infty 2^{n}e_n$ and likewise for $f$. Then, one has $$x = y_0 \le \sum_n 4^{n}e_n\otimes f_n \le e\otimes f.$$ If $A$ and $B$ have strictly positive elements, one can arrange $(e\otimes f)  x$ is strictly positive.
I think with more efforts one can find $e$ and $f$ such that $\ e \ \ f \ = \ x \$.

$\begingroup$ Thank you, But could you please explain why the last inequality hold? $\endgroup$ – Ali Taghavi Jul 11 '13 at 11:03

$\begingroup$ That's because $e \otimes f = \sum_{m,n}2^{(m+n)}e_m \otimes f_n$ is an unconditionally norm convergent series of positive elements. $\endgroup$ – Narutaka OZAWA Jul 11 '13 at 23:38